Asking for some hints for this real analysis question

Question

Justify the following equation:

$$\int_0^1 \frac{x^{p-1}}{1-x}\log(\frac{1}{x}) dx=\sum_{n=0}^{\infty} \frac{1}{(n+p)^2} $$ Where $p>0$

Answer 1

Hint:

1. $\frac{1}{1-x}=\sum_{n=0}^\infty x^n$.
2. Motivate $\int\sum=\sum\int$.
3. At last, integration by parts.

Answer 2

Hint. By using a classic geometric series result, one has $$ \frac{x^{p-1}}{1-x}=\sum_{n=0}^\infty x^{n+p-1},\quad |x|<1, $$ then observing that $$ \int_0^1x^{a-1}\ln x\:dx= -\frac1{a^2} $$ one gets the expected result with a termwise integration.